Let $P_1P_2\cdots P_n$ be a convex polygon (two consecutive points are distinct and three vertices can't be in the same line).
Is it possible that there exists $\vec{v}\neq \vec{0}$, $\alpha>0$, $\beta >0$ and $i \neq j$ such that $\vec{P_iP_{i+1}} = \alpha \vec{v}$ and $\vec{P_jP_{j+1}} = \beta \vec{v}$ ?
I think that this answer is no but I don't know how to prove it.
Take a moment to convince yourself that the sum of the exterior angles of any polygon is $2 \pi$, where we define the exterior angle $\angle V_k$ as the rotation angle measured counterclockwise at a vertex $V_k$ from the entering segment $\vec{P_{k - 1} P_k}$ to the exiting segment $\vec{P_k P_{k + 1}}$, where the interior of the polygon is always on the left side of these segments (we traverse the vertices in a "counterclockwise" fashion). Clockwise angles are taken to be negative; i.e., each exterior angle satisfies $\angle V_k \in (-\pi, 0) \cup (0, \pi)$. The values $-\pi, 0, \pi$ are not included in this range since we do not consider a straight segment as multiple sides.
Next, notice that the situation in which two distinct sides $\vec{P_i P_{i + 1}}$ and $\vec{P_j P_{j + 1}}$ ($i < j$) are parallel and oriented in the same direction means that the sum of the exterior angles between these two sides obeys exactly one of the following:
$$\begin{cases} \sum_{k \in \{i + 1, ..., j\}} \angle V_k = 0 && \text{and} && \sum_{k \in \{1, ..., i\} \cup \{j + 1, ..., n\}} \angle V_k = 2 \pi && \text{or} \\ \sum_{k \in \{i + 1, ..., j\}} \angle V_k = 2 \pi && \text{and} && \sum_{k \in \{1, ..., i\} \cup \{j + 1, ..., n\}} \angle V_k = 0 \end{cases}$$
That is, there exists some set of consecutive vertices whose exterior angles sum to $0$. Note that this set cannot be empty, because that would imply that $i = j$. This set also cannot contain just $1$ element, because that would imply that for some $k \in \{1, ..., n\}$, $\angle V_k = 0$. So, a set of at least two exterior angles sums to $0$, which implies that at least one $\angle V_k < 0$, and this will be a reflexive angle. Therefore, the polygon cannot be convex.